13 CHEM 140S: Expt 12 Reactions in Aqueous Solutions Post-lab Questions 1. The f

Question

13 CHEM 140S: Expt 12 Reactions in Aqueous Solutions Post-lab Questions 1. The foll when added owing salts are soluble in water. Write the equation for the dissociation of each of the salts to water. One example is provided. eg sodium sulfate. Na2SO4(s) SO42-tao + 2Na++ aq 1405 a) Lead(lI) nitrate b) Iron(II) chloride is the c) Ammonium phosphate d) Potassium carbonate 2. W hat is the concentration of the magnesium ions and the chloride ions in 2.5 M solution of MgCl2? 3 (a) Suppose l 00 mL of 0.70 % glucose solution was diluted to 10.00 mL (by adding water) what is the concentration of the diluted solution? (b) If 1.00 mL of the diluted solution in step (a) was further diluted to 10.00 mL (by adding water), what is the concentration of the solution after the 2nd dilution?

Explanation / Answer

1)

a) Pb(NO3)2 (s) + aq ------> Pb2+(aq) + 2NO3-(aq)

b) FeCl2 (s)   + aq -------> Fe2+ (aq) + 2Cl- (aq)

c) (NH4)3PO4(s) + aq ------> 3NH4-(aq) + PO43-(aq)

d) K2CO3(s) + aq -------> 2K+(aq) + CO32-(aq)

2)

MgCl2 is strong electrolyte, it is completely dissociate into Mg2+ and Cl-

  MgCl2 --------> Mg2+    + 2Cl-

1 mole of MgCl2 give 1mole of Mg2+ and 2mole of Cl-

Concentration of MgCl2 = 2.40M

Therefore,

Concentration of Mg2+ = 2.40M

Concentration of Cl-= 2× 2.40M = 4.80M

3)

a)Formula is

C1 × V1 = C2 × V2

C1 = initial concentration,0.70%

V1 = Initial volume , 1ml

V2 = Final volume , 10ml

C2 = Final Concentration

C2 = V1×C1/V2

= 1ml × 0.70%/10ml

= 0.07%

So, the diluted concentration is 0.07%

b) Apply the formule in a)

C2 = C1 × V1/V2

= 0.07% × 1ml /10ml

= 0.007

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